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A disk on a slippy surface (ROTATION &TRANSLATION)

  1. Jan 27, 2017 #1
    1. The problem statement, all variables and given/known data
    There is a disk with mass 50 grams put on a slippy surface (no friction!). Its radius is 5 cm. Mass of the weight is 20 grams. At the beginning there is no motion. How far does the center of mass of the disk move in 10 seconds? (answer: 91 m) How many turns does the disk in 10 seconds?
    Capture.jpg

    2. Relevant equations
    F=ma
    F*r = J*alpha

    3. The attempt at a solution
    The only force on the disk is the force of the rope if there is no friction. -->

    Capture1.jpg

    I did not solved the second question as the first answer is wrong. Please help me, correct me.
     
  2. jcsd
  3. Jan 27, 2017 #2

    TSny

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    Is the acceleration of the center of the disk the same as the acceleration of the hanging weight?
     
  4. Jan 27, 2017 #3
    Yeah, right, they are not. But I do not know how to solve it in other way
     
  5. Jan 27, 2017 #4
    acceleration of the disk = force of the rope / mass of the disk ... Is that correct?
     
  6. Jan 27, 2017 #5

    TSny

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    Yes. ("Acceleration of the disk" means the acceleration of the center of the disk.)
     
  7. Jan 27, 2017 #6
    Yes, I meant that
     
  8. Jan 27, 2017 #7

    TSny

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    You will need to relate the acceleration of the hanging weight to the acceleration of the center of the disk and the angular acceleration of the disk.
     
  9. Jan 27, 2017 #8
    angular acceleration (alpha) = (2 * force_rope )/ (mass * radius)
     
  10. Jan 27, 2017 #9
    yeah,yeah, i think i got it! circimferential acceleration + acceleration of center of mass = acceleration of weight , right?
     
  11. Jan 27, 2017 #10
    Ok, thanks, I got the right result! Your hints are very useful! Can you give me your phonenumber? ... just joking! have a nice day, thanks!
     
  12. Jan 27, 2017 #11

    TSny

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    Good work!
     
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